Mechanics Tutorial 6 - Motion in a Straight Line

Speed, Velocity, and Acceleration

This topic looks at linear motion, i.e. motion in a straight line.

  You will be familiar with the simple equations:


            speed (m/s) = distance (m)        



            velocity (m/s) = displacement (m)

                                                time (s)


In Physics code this is written:

We also know acceleration as:

acceleration (m s-2) = change in velocity (m s-1) ÷ time (s)

In Physics code:

In both these equations, the strange triangular symbol, D, is Delta, a Greek capital letter 'D', which is the physics code for "change in".

The picture below shows the difference between distance and displacement.

Suppose we have two towns A and B 10 km apart but either side of a hill.  They are joined by a railway line that is straight, and goes through the hill in a tunnel.  The road goes round the hill and the total journey distance is 25 km.

So the distance is 25 km.  The displacement (the straight-line distance in a particular direction) between A and B is 10 km due East.

If we go from A to B and back again, the distance is 50 km, but the displacement is 0.

Question 1

Why is the displacement 0?


Question 2

A runner accelerates at a rate of 4 ms-2 to her maximum speed of 9.6 m/s.  What is the time taken for her to reach this speed?


Information and Communication Technology (Computers) can be used to demonstrate the motion of a vehicle.  Sensors (light gates or ultrasonic detectors) are connected to a computer and the computer will record the data at time intervals.  The computer will plot a graph of the motion.

Here are some important definitions.


Note that some textbooks (and examination boards) use "velocity" as a posh word for speed.  It has a specific meaning, displacement divided by time.



Graphical Interpretation of Acceleration

We can represent the movement of objects using a graph, usually plotting time on the x-axis (horizontal) and the speed or distance on the y-axis (vertical). 

Here is a displacement-time graph:


For a distance time graph, the gradient would be the speed.

Consider a train accelerating from a station along a straight and level track to a maximum speed and slowing down to a stop at the next station.  The easiest way to show this is with a speed time graph.

Acceleration is the gradient of the speed-time graph.

From the graph,

Distance is the area under the speed-time graph.  To work out the total distance, we would add the areas of:

Question 3

Use the information below in the question.  The train's motion is that described in the graph above.

  • The maximum speed of the train is 25 m/s

  • The time interval OX is 45 s

  • The time interval XY is 45 s

  • The time interval YC is 20 s         

What is:

a)      The acceleration between O and A

b)      The acceleration between B and C

c)      The distance covered while the train is at constant speed

d)      The total distance.

    e)      The average speed?  


The corresponding distance time graph is like this:


We can work out the speed at any instant by measuring the gradient of the distance time graph. The curved line tells us that the speed is changing.

Acceleration is usually uniform, which means that the speed [velocity] is changing at a constant rate.    This is shown by a straight line on a speed time graph.  However in many real life situations acceleration is not constant.  Therefore the graph is not a straight line:

Equations of Motion

We can use the Equations of Motion to calculate the speed of an object under different circumstances.  These are quantities are involved in linear motion, movement in a straight line:


Physics Code





Speed at the start



Speed at the end









1.  Speed at finish = speed at start + change in speed

change in speed = acceleration × time.

Speed at end = speed at start + (acceleration × time)


3.  Distance = average speed × time




Question 4

A car is travelling at 30 m/s and takes 10 seconds to acceleration to a new speed of 35 m/s.  What is its acceleration? 


Question 5

A brick falls off the top of a wall under construction and drops into a bed of sand 14.5 m below.  It makes a dent in the sand 185 mm deep.  What is:

a)       The speed of the brick just before it hits the sand.                                                         

b)       Its deceleration in the sand.                                                                                          

c)       What would happen to a person undergoing that deceleration? 



In many examples we can ignore air resistance, although you will know for yourselves that the faster you go on a mountain bike, the harder you have to pedal.  This is because of the effects of friction and air resistance (drag).  We will look at this next in Terminal Velocity.  


How are the equations of motion derived?

All the equations of motion are derived from the speed (or velocity) time graph. 


1. The first one is quite easy as it's derived from the area under the graph:


2.  This graph shows acceleration:


3. This derivation is a little more complex to understand:


4. The final derivation requires some manipulation of equations: 


How do we select the right equation?

The flow chart below can be used to help you to select the correct equation of motion to use:




Question 6

An aeroplane of mass 5000 kg lands on a runway at a speed of 60 m/s and stops 25 s later.




a – the deceleration of the aeroplane;

b – the braking force on the aeroplane.

c - the distance taken by the aeroplane to stop.